Question

If a, b, c, d are in continued proportion, prove that: `sqrt(ab) + sqrt(bc) - sqrt(cd) = sqrt((a + b - c)(b + c - d))`.

Answer 1

a, b, c, d are in continued proportion,

`a/b = b/c = c/d` = k

c = dk, b = ck = dk², a = bk = dk³

L.H.S.

= `sqrt(ab) + sqrt(bc) - sqrt(cd)`

= `sqrt((dk^3)(dk^2))  + sqrt((dk^2)(dk)) - sqrt((dk)(d))`

= `sqrt(d^2k^5) + sqrt(d^2k^3) - sqrt(d^2k)`

= `dk^2 sqrt(k) + dk sqrt(k) - dsqrt(k)`

Factoring out `d sqrt(k)`

= `d sqrt(k) (k^2 + k - 1)`

R.H.S.

First, simplify the terms inside the parenthesis:

a + b − c

= dk³ + dk² − dk

= dk(k² + k − 1)

b + c − d

= dk² + dk − d

= d(k² + k − 1)

Now substitute these into the RHS:

= `sqrt((a + b - c)(b + c - d))`

= `sqrt([dk(k^2 + k - 1)][d(k^2 + k - 1)])`

= `sqrt(d^2k(k^2 + k - 1)^2)`

= `d sqrt(k) (k^2 + k - 1)`

∴ L.H.S. = R.H.S.

Hence proved.